1. Sep 18, 2007

### tigerd12

1. The problem statement, all variables and given/known data

Hello, I just found a question, and having attempted it many times I get different answers, probably due to my messy working, however I have just tried it twice again and got the same answer, just checking with you guys to see if you think it is correct.

Thanks!

Simplify fully

2/(x-1)+(x-11)/(x^2+3x-4)

Hope it's right!

Cheers.

2. Sep 18, 2007

### neutrino

Is this the fraction $$\frac{2}{x-1} + \frac{x-11}{x^2+3x-4}$$? If so, you have missed a factor of 3.

3. Sep 18, 2007

### tigerd12

Post

Hmm?

What do you mean?

yes the question is as you stated

4. Sep 18, 2007

### tigerd12

I started out solving the question by getting lowest common denominator.. i.e

(x-1)(x+4)+(x-1) was my bottom line.. and then i carried on.. so which factor did I miss?

5. Sep 18, 2007

### neutrino

What did you do after that?

6. Sep 18, 2007

### tigerd12

Working

Ok here is the working..

2(x-1)(x+4)+(x-1)(x-11)/(x-1)(x+4)+(x-1)

Simplifies into..

2x^2+6x-8+x^2-11x-x+11/(x-1)(x+4)(x-1)

which then is

3x^2-6x+3/(x-1)(x+4)(x-1)

which then is

x^2-2x+1/(x-1)(x+4)(x-1)

which then turns out to be

(x-1)(x-1)/(x-1)(x+4)(x-1) = 1/(x+4)

I think..

7. Sep 18, 2007

### neutrino

Hmm...I'm not sure how you arrived at that, but

$$\frac{2}{x-1} + \frac{x-11}{x^2+3x-4}$$ = $$\frac{2(x+4) +(x-11)}{(x-1)(x+4)}$$

Think of it as multiplying the numerator and denominator of the first term by (x+4) and simplifying.

8. Sep 18, 2007

### tigerd12

ahhhhh i get you, i understand, didnt spot that (x-1) was also factor or the other quadratic.. but 1 thing..

the way I got to my answer is taking the LCM by multiplying the two bottom terms, then working from there.. theoretically it should have come out right, i can't see why it didn't

9. Sep 18, 2007

### neutrino

The LCM of (x-1) and (x-1)(x+4) is (x-1)(x+4). Therefore the denominator becomes (x-1)(x+4), and you multiply the 2 by (x+4) and the (x-11) by 1.

10. Sep 18, 2007

### tigerd12

yeah I get that, but couldn't (x-1)(x-1)(x+4) be used too as its still a multiple of both

11. Sep 18, 2007

### neutrino

We're talking about least common multiple.

12. Sep 18, 2007

### tigerd12

I thought when adding fractions you just needed a common multiple, not the lowest? or am I wrong?

13. Sep 18, 2007

### neutrino

You said you took the LCM. What do you think LCM stands for?

Here's a numerical example: (1/2) + (1/4) = 0.50 + 0.25 = 0.75

The least common multiple of 2 and 4 is 4, since 2x2 = 4x1 = 4.

(1/2) + (1/4) = ((1x2) + (1x1))/4 = 3/4 = 0.75

Last edited: Sep 18, 2007
14. Sep 18, 2007

### tigerd12

Yeah, I thought I took what I thought was the LCM, I was wrong, but what i'm asking is that shouldn't it have worked as the denominator I gave was still a multiple of both

15. Sep 18, 2007

### neutrino

Ah. What you essentially did was multiply the numerator and denominator by (x-1). So the answer should be the same as the correct one.

But from going from 3x^2-6x+3/(x-1)(x+4)(x-1) to x^2-2x+1/(x-1)(x+4)(x-1), you forgot the factor of three in the numerator, just as I said in my first post.

16. Sep 18, 2007

### tigerd12

But doesn't 3x^2-6x+3 = x^2-2x+1? As I took out the factor of 3?

17. Sep 18, 2007

### neutrino

But where does the 3 go?

What you're saying is the same as stating 3 = 1. Does that make sense?

3x^2-6x+3 is equal to thrice x^2-2x+1.

Since 3 is common to all terms, I can write it as 3(x^2-2x+1). Multiply each term within the brackets by 3 and you'll get back to old expression.

18. Sep 18, 2007

### tigerd12

hmm ok.. kinda wierd, since i thought dividing everything by three should keep it the same